SME
Circuit Analysis I · Problem type
Norton equivalent
Use this for “Turn an 8 V, 2 Ω Thévenin equivalent into its Norton form.”
Diagram
IN
=
VTh
RTh
=
8
V
2
Ω
=
4
A
The same source seen from a and b: either form gives the load the same V and I.
This is an example. Type your own values to start a new problem.
VTh (V)
=
IN (A)
×
RTh (Ω)
Formulas
VTh = IN × RTh
8 = 4 × 2
Assumptions
•
The Norton form is a current source IN in parallel with RN = RTh.
•
IN is the short-circuit current at a–b; VTh is the open-circuit voltage there.
•
Both forms give any load the same voltage and current.
Step-by-step
WE KNOW
VTh = 8 V, RTh = 2 Ω
FIND
Norton current (IN)
Step 1 · Find Norton current (IN)
Use VTh = IN × RTh
Short
a–b:
the
Thévenin
source
drives
VTh
RTh
through
the
short.
IN
=
VTh
RTh
IN
=
8
2
IN = 4 A
Check
Put the numbers back in. Both sides should match.
8 = 4 × 2 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Thévenin and Norton equivalents
Main lesson
Maximum power transfer
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Use this for “A 12 V source through 3 Ω and a 3 A current source feed one node, with 6 Ω to ground. Find the node voltage by superposition.”